SOLUTION: A teacher can sell 123 of her pies at $4 per pie. for each $0.50 decreases in price, she can sell 20 additional pies the quadratic equation = -10x^2+18.5x+492 what price shou

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A teacher can sell 123 of her pies at $4 per pie. for each $0.50 decreases in price, she can sell 20 additional pies the quadratic equation = -10x^2+18.5x+492 what price shou      Log On


   



Question 1159587: A teacher can sell 123 of her pies at $4 per pie. for each $0.50 decreases in price, she can sell 20 additional pies
the quadratic equation = -10x^2+18.5x+492
what price should she charge to maximize her revenue
how many pies does she need to sell in order to ensure a revenue of $450

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A teacher can sell 123 of her pies at $4 per pie. for each $0.50 decreases in price, she can sell 20 additional pies
the quadratic equation = -10x^2+18.5x+492
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An equation states that 2 sides are equal.
= -10x^2+18.5x+492 has nothing on the Left Side.
---> not an equation
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Yes, I do mean to be "picky." It's math. It's not a guessing game.
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what price should she charge to maximize her revenue
how many pies does she need to sell in order to ensure a revenue of $450